Finding the expected value $E(X)$ and $E(Y)$

Question

A random variable $N$ is uniformly distributed on $\{1,2,...,10\}$ Let $X$ be the indicator of the event($N\le 5$) and $Y$ be the indicator of the event ($N$ is even)

So I have to find $E(X)$ and $E(Y)$

From the formula sheet I know

$$E(X) = \sum_i x_iP(x_i)$$ $$E(Y) = E[E(Y\mid X=x_i)] $$

but i don't get how to select the xi and P(xi) to solve this problem. The answer says $E(X)=E(Y) =1/2$ but It does not have any process inside so I could not understand can anyone help me about this

Answer 1

Looking at RHS of equality: $$\mathbb EX=\sum_ix_iP(x_i)$$ we observe that a term $x_iP(x_i)$ is only relevant if: $$x_iP(x_i)\neq0$$

That will be the case here if and only if $x_i=1$ and in that case: $$P(x_i)=P(1)=\Pr(X=1)=\Pr(N\leq5)=\frac5{10}=\frac12$$

So we find:$$\mathbb EX=1\cdot\frac12=\frac12$$You can find $\mathbb EY$ on a similar way.

Answer 2

The two rv's are both bernulli with parameter 0.5 thus their expectation is $0\times 0.5 + 1\times 0.5 =0.5$

This because the rv that describes the event $N\leq 5$ can take only the values 0 and 1 (false or true) with probability 0.5 as you have 5 favourable events among 10 equiprobable ones

Similar reasoning for the other rv